Problem
Let $\triangle ABC$ be a triangle, and let $P$ and $Q$ be points on $\overline{AB}$ and $\overline{AC}$ respectively, such that $\frac{AP}{AB} = \frac{AQ}{AC}$. Prove $\overline{PQ} \parallel \overline{BC}$.
Attempt
The relationship $\frac{AP}{AB} = \frac{AQ}{AC}$ implies $\frac{AP}{PB} = \frac{AQ}{QC}$. And this implies that the shortest distance from $P$ to $\overline{BC}$ is equal to the shortest distance from $Q$ to $\overline{BC}$.
The Lemma on Shortest Distance states that if the shortest distance between $P$ and $\overline{BC}$ is the same as the shortest distance between $Q$ and $\overline{BC}$, then $\overline{PQ} \parallel \overline{BC}$. Thus, $\overline{PQ} \parallel \overline{BC}$.
Question
I don't think this is an adequate explanation for why the shortest distance from $P$ to $\overline{BC}$ is equal to the shortest distance from $Q$ to $\overline{BC}$. In short, why is that implied by $\frac{AP}{AB} = \frac{AQ}{AC}$?